Answer to Question #229223 in Chemical Engineering for Lokika

Question

Solve dy/dx= 4x² (y-x)² + if y /x is a particular solution.

Answer 1

Given DE:

$$dy/dx= 4x² (y-x)² +y/x$$

Let

$$u(x)=y/x$$

Then

$$y=u\cdot x,\quad y'=u'x+u$$

We get

$$u'x+u=4x^2(ux-x)^2+u$$ $$u'x=4x^4(u-1)^2$$

$$\frac{du}{(u-1)^2}=4x^3dx$$

$$-\frac{1}{u-1}=x^4+C$$

$$u-1=-\frac{1}{x^4+C}$$

$$u=1-\frac{1}{x^4+C}$$

Finally

$$y(x)=x\left(1-\frac{1}{x^4+C}\right)$$